Question

A radioactive element A decays by the sequence and with half-lives given below:
$A\xrightarrow[30\min ]{\alpha }B\xrightarrow[2days]{2\beta }C$
Which of the following statements about this system are correct?
(A) The mass number B is greater than A.
(B) After two hours, less than 10% of the initial A is left.
(C) Maximum amount of B present at any time is less than 50% of the initial amount of A.
(D) The atomic numbers of A and C are the same.

Answer 1

The atoms which are unstable and emit radiation to attain stability are called radioactive elements. After emitting the radiation the radioactive element changes into another element which may be stable. The process of emitting radiations is called decay.

Complete step by step answer:
The given chemical equation is as follows.
$A\xrightarrow[30\min ]{\alpha }B\xrightarrow[2days]{2\beta }C$
Here the radioactive element in the given equation is A.
As per the above chemical equation, A is a radioactive element and undergoes alpha ($\alpha$) decay and forms B is the product in 30min.
The formed product B is also unstable and undergoes decay and liberates two beta ($\beta$) particles and forms C as a product.
Means totally there is an emission of one alpha and two beta particles to reach product C.
We can see the emission of alpha and beta rays as follows.
${}_{Z}^{M}A\xrightarrow[30\min ]{\alpha }{}_{Z-2}^{M-4}B\xrightarrow[2days]{2\beta }{}_{Z-2+2}^{M-4}C\to {}_{Z}^{M-4}C$
Here Z = atomic number, M= Mass number of the elements.
When an atom loses an alpha particle then the atomic number of the atom decreases by two units and mass number decreases by four units.
When an atom loses two beta particles then the atomic number of the atom increases by two units.
If an atom loses one alpha and two beta particles the resulting element will be an isotope of the original atom.
Isotopes have the same atomic number and different mass number.
Therefore, the atomic numbers of A and C are the same.

So, the correct option is D.

Note: -Alpha particle ($\alpha$) is going to represent as ${}_{2}^{4}He$.
-The mass number of the alpha particle is 4 and the atomic number is 2.
-Beta particle is going to represent as${}_{-1}^{0}\beta$.
-The mass number of the beta particle is 0 and the atomic number is -1.